Do extension fields always belong to a bigger field? Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?
Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?
 A: You could embed $F$ into its algebraic closure $\overline{F}$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $\overline{F}$ and we can take the minimal subfield of $\overline{F}$ that contains $E_1 \cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.
A: Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product
$A=E_1\otimes_F E_2$ is a commutative ring, not necessarily a field though.
Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument.
Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $x\mapsto
\overline{x\otimes 1}\in A/I$ is a ring homomorphism $E_1\to K$. As $E_1$
is a field, this is an injective homomorphism, so we can think of $E_1$
being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.
A: For your specific example:
Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=\prod_{i=1}^{\ell} q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $\prod_{i=2}^{\ell} q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.
